Question

9. Empty taxis pass by a street corner at a Poisson rate of two per minute and pick up a passenger if one is waiting there. Passengers arrive at the street corner at a Poisson rate of one per minute and wait for a taxi only if there are less than four persons waiting; otherwise they leave and never return. Ella arrives at the street corner at a given time. Find her expected waiting time, given that she joins the queue. Assume that the process is in steady-state.

Answer 1

0.37

Explanation:

To resolve the given problem we will apply

persistent time Markov chain : n = 0, 1, . . . , 4

n = number of individuals pausing

where The equalization conditions are : πn = ( πn - 1/2 )

Given that :

n = 0,1,2,3,4

π0 = 1/( 1 + 2−1 + 2−2 + 2−3 + 2−4) = 0.52 = 16/31

Also the normal number of travelers found by Ella will be represented as

E(N) = (π1 + 2π2 + 3π3) / ( π0 + π1 + π2 + π3) ------- ( 1 )

where : π1 = 8, π2 = 4, π3 = 2 , π0 = 16/31 input values into equation 1

E ( N ) = 22/30

given that the True to form hanging tight time = 0.5

hence Holding holding time = E(N ) * 0.5

therefore the expected waiting time for Ella = ( 22/30 ) * 0.5 = 0.37